Problem 29 - Precalculus Essentials

Question

Involve equations with multiple angles. Solve each equation on the interval \([0,2 \pi)\) $$\tan 3 x=\frac{\sqrt{3}}{3}$$

Step-by-Step Solution

Answer

The angles which satisfy the equation \(\tan 3x = \frac{\sqrt{3}}{3}\) within the interval \([0,2\pi)\) are \(x = \frac{\pi}{18}, \frac{7\pi}{18}, \frac{13\pi}{18}, \frac{19\pi}{18}, \frac{25\pi}{18}, \frac{31\pi}{18}, \frac{37\pi}{18}, \frac{43\pi}{18}, \frac{49\pi}{18}, \frac{55\pi}{18}, \frac{61\pi}{18}\)

Step 1: Identify the nature of the equation

Identify that \(\tan 3x = \frac{\sqrt{3}}{3}\) is a trigonometric equation, implicating use of the unit circle and related angles. Furthermore, acknowledge that this equation has a multiple angle \(3x\).

Step 2: Recognize the values on the unit circle

Realize that \(\frac{\sqrt{3}}{3}\) is equal to \(\tan(\frac{\pi}{6})\), therefore the equation simplifies to \(\tan(3x) = \tan(\frac{\pi}{6})\). Here, \(3x\) and \(\frac{\pi}{6}\) are angles corresponding to the same tangent value.

Step 3: Get the general solution

For tangent, the function is periodic with period \(\pi\), hence the solution can be represented in the form: \(3x = \frac{\pi}{6} + k\pi\), where \(k\) is an integer.

Step 4: Solve for x

Divide the general solution by \(3\) to solve for \(x\): \(x = \frac{\pi}{18} + \frac{k\pi}{3}\), where \(k\) is an integer.

Step 5: Find all the solutions within the given interval

Substitute different integer values for \(k\) to find \(x\) that satisfies the interval \([0,2\pi)\). The solutions are \(x = \frac{\pi}{18}, \frac{7\pi}{18}, \frac{13\pi}{18}, \frac{19\pi}{18}, \frac{25\pi}{18}, \frac{31\pi}{18}, \frac{37\pi}{18}, \frac{43\pi}{18}, \frac{49\pi}{18}, \frac{55\pi}{18}, \frac{61\pi}{18}\)

Key Concepts

Understanding the Unit Circle

The unit circle is a key tool in trigonometry that helps in understanding the relationship between angles and trigonometric functions like sine, cosine, and tangent. It is a circle with a radius of one centered at the origin of a coordinate plane. Each point on the circumference of the unit circle represents the cosine and sine of an angle measured from the positive x-axis.
Using the unit circle, we can easily find the value of trigonometric functions for angles that overlap standard positions like \(0\), \(\frac{\pi}{6}\), \(\frac{\pi}{3}\), \(\frac{\pi}{2}\), and more.

• For example, at \(\frac{\pi}{6}\) radians, the coordinates are \((\frac{\sqrt{3}}{2}, \frac{1}{2})\), making the tangent (sine divided by cosine) equal to \(\frac{1}{\sqrt{3}}\) or \(\frac{\sqrt{3}}{3}\).

This understanding allows us to quickly determine that \(\tan(3x) = \frac{\sqrt{3}}{3}\) corresponds to \(\tan(\frac{\pi}{6})\) on the unit circle, affirming the critical importance of the unit circle in solving trigonometric equations.

Exploring Multiple Angle Identities

Multiple angle identities involve trigonometric equations where the angle has a coefficient greater than one, such as \(3x\) in \(\tan(3x) = \frac{\sqrt{3}}{3}\). These identities help us rewrite expressions like \(\sin(2x)\), \(\cos(3x)\), or \(\tan(nx)\), making complex equations more manageable.
The crucial aspect here is to acknowledge that such equations expand the number of potential solutions because of the nature of trigonometric functions being periodic. The general formula for the tangent function's periodicity is given by:

• \(\tan(\theta + k\pi) = \tan(\theta)\) for any integer \(k\).

In our exercise, because \(3x\) acts as the angle within the tangent function, the solutions will repeat every \(\pi\) radians. By solving \(3x = \frac{\pi}{6} + k\pi\), we can find the general solution and, subsequently, each specific solution within the prescribed interval.

Finding General Solutions in Trigonometry

General solutions in trigonometry are expressions that describe all the potential values of \(x\) that satisfy a trigonometric equation across all possible cycles of the function. The periodicity of trigonometric functions involves knowing:

• Sine and cosine have a period of \(2\pi\).
• Tangent has a period of \(\pi\).

When solving for \(x\) in \(\tan(3x) = \frac{\sqrt{3}}{3}\) over \([0,2\pi)\), the general solution \(3x = \frac{\pi}{6} + k\pi\) is derived using the periodicity of the tangent function.
To isolate \(x\), you must solve \(x = \frac{\pi}{18} + \frac{k\pi}{3}\) for integer values of \(k\). By systematically substituting integer values for \(k\), all solutions within the appropriate range can be determined. This method ensures that every valid angle within the \([0,2\pi)\) range is accounted for by the general solution.